Exploring Heisenberg's Uncertainty Principle: Intuition & Explanations

[jeebs]

You are no doubt familiar with Heisenberg's uncertainty principle, putting a limit on the accuracy with which we can measure a particle's position and momentum, $$\Delta x \Delta p \geq \hbar/2$$
On my course I was shown the derivation, it popped out of a few lines of mathematics involving the Cauchy-Riemann inequality.

However, I've been wondering if there is any reason to intuitively expect difficulties when trying to simultaneously know both quantities. What I mean is, is there anything about the nature of "position" and "momentum" that hints that we should not be able to know both simultaneously?

One explanation I heard was that if you, say, bounced a photon off an atom to measure its position, then the recoil would affect its momentum, thus giving rise to the uncertainty - this seems straightforward enough. However, I have also been told that this is apparently not a valid explanation, although I do not understand why.
Can anyone shed any light on this for me?

 

[CJames]

The intuitive explanation you present is called Heisenberg's microscope. There are two primary problems with it. The first is that it only results in an approximate expression of the equation that you cite.

The second is that it attacks its own premises. The thought experiment first assumes that the electron has a definite location and momentum, and then demonstrates why such a thing can't exist, which invalidates its own premises.

 

[nkadambi]

Here is a very general answer: From the axioms of QM and the math that is used to build observables and states of systems, it turns out that position and velocity (and also momentum, because momentum p = mv) are what are called "canonical conjugates", and they cannot be both be "sharply localized". That is, we cannot measure them both to an arbitrary level of precision. It is a mathematical fact that any function and its Fourier transform cannot both be made sharp.

This is a purely a mathematical fact and so has nothing to do with our ability to do experiments or our present-day technology. As long as QM is based on the present mathematical theory, it cannot be done using the mathematics we have.

 

[eaglelake]

There is nothing in Classical theory that prevents us from knowing both the momentum and the position of a particle with "certainty". i.e. we can repeat the classical experiment many times and always get the same result for momentum and the same result for position.

But in quantum mechanics, position and momentum are linear operators in a Hilbert space and, most importantly, the momentum operator and the position operator do not commute. This means that there is no wavefunction that is a common eigenfunction of both momentum and position. Further, we must know the wavefunction in order to calculate the uncertainties. For practice, make up a (simple) wavefunction and do the caculations for $$\Delta x$$ and $$\Delta p$$ and then take their product to convince yourself that the uncertainty principle is satisfied. I am trying to emphasize that this is quantum mechanics and not classical physics.

Bouncing a photon off an atom tells us nothing about any uncertainties. We must bounce many identically prepared photons off like atoms in order to get the statistical distributions of atomic position measurements and atomic momentum measurements. What we call "uncertainty" is a property of a statistical distribution. You cannot determine an uncertainty from a single measurement. I hope this helps.
Best wishes.

 

[Fredrik]

Good post nkadambi, but I must point out an inaccuracy in what you said. I didn't really understand this myself until recently. It is possible to measure position and momentum simultaneously. In fact, we often measure the momentum by measuring the position and interpreting the result as a momentum measurement. (Check out figure 3 in this pdf). What we can't do is to prepare a state such that we would be able to make an accurate prediction about what the result of a position measurement would be and an accurate prediction about what the result of a momentum measurement would be.

jeebs: Mathematically, the "uncertainty" is derived from the axioms of QM, and is only non-zero if the commutator of the two operators is non-zero. Physically, I think the problem is always that a device that prepares a state with a sharply defined value of one of the observables would interfere with a device that prepares a state with a sharply defined value of the other observable. So... non-zero uncertainty = non-commutativity = the state preparation devices would interfere with each other.

 

[nkadambi]

Thanks for the post, Fredrik. And thanks for the paper: I'll check it out.

I am new to this forum (I just registered couple hours ago!) I mainly have a pure math background, and only just starting into mathematical physics. Are you a grad student or professor? At my school there is hardly anyone who ventures into math physics. I am looking to make a few friends online with similar background.

 

[CJames]

Wow Fredrik, that is incredibly helpful. Just when I thought I was starting to understand something...

 

[dlgoff]

It is possible to measure position and momentum simultaneously.

I've always liked how ZapperZ explains this in is blog http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html" .

 

[Fredrik]

I am new to this forum (I just registered couple hours ago!) I mainly have a pure math background, and only just starting into mathematical physics. Are you a grad student or professor? At my school there is hardly anyone who ventures into math physics. I am looking to make a few friends online with similar background.

You have come to the right place. There are plenty of people here with all sorts of backgrounds.

I've always liked how ZapperZ explains this in is blog http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html" .

Yes, ZZ understood this a long time before I did. I would have understood it much sooner if I had read his posts in these threads more carefully. I was naive enough to think I already understood these things.

 

[Naty1]

(sorry this is so long but I have just been struggling through the same concepts.)

I hope the essence of Zapper's HUP explanation is here:

The HUP isn't about a single measurement and what can be obtained out of that single measurement. It is about how well we can predict subsequent measurements given the identical conditions.

and

What I am trying to get across is that the HUP isn't about the knowledge of the conjugate observables of a single particle in a single measurement. I have shown that there's nothing to prevent anyone from knowing both the position and momentum of a particle in a single mesurement with arbitrary accuracy that is limited only by our technology. However, physics involves the ability to make a dynamical model that allows us to predict when and where things are going to occur in the future. While classical mechanics does not prohibit us from making as accurate of a prediction as we want, QM does!

Somebody in the recent past posted this...my boldface.. (I did not record the poster, maybe even Zapper??..was a trusted source here.) I'm posting this to confirm that it is an equivalent description, that it matches Zappers blog...

...to measure a particle's momentum, we need to interact it with a detector, which localizes the particle. So we actually do a position measurement (to arbitrary precision). Then we calculate the momentum, which requires that we know something else about the position of the particle at an earlier time (perhaps we passed it through a narrow slit). Both of those position measurements, and the measurement of the time interval, can be done to arbitrary precision, so we can calculate the momentum to arbitrary precision. From this you can see that in principle, there is no limitation on how precisely we can measure the momentum and position of a single particle.

Where the HUP comes into play is that if you then repeat the same sequence of arbitrarily precise measurements on a large numbers of identically prepared particles (i.e. particles with the same wave function, or equivalently particles sampled from the same probability distribution), you will find that your momentum measurements are not all identical, but rather form a probability distribution of possible values for the momentum. The width of this measured momentum distribution for many particles is what is limited by the HUP. In other words, the HUP says that the product of the widths of your measured momentum probability distribution, and the position probability distribution associated with your initial wave function, can be no smaller than Planck's constant divided by 4 times pi

So what I think these mean is that you can get precise but not necessarily ACCURATE simultaneous measurements...that is, you cannot REPEAT the exact measurement results as is possible to arbitrary precision in classical measurements. What had me confused, and I hope I understand better, was that commutativity and non commutativity of operators applies to the distribution of results, not an individual measurement.

In Quantum Mechanics, Albert Messiah provides an interpretation for the inability to repeat the measurements :

Immediately after the operation of a measurement the system is in a dynamical state with the arbitrarily precise position measured; Such a state cannot be represented by a wave function (psi). The state psi in general corresponds to a probability distribution of finding some value x, not a precise value of x…..or of measuring momentum. The function psi does not give more than the statistics of positions…or momenta... During the process of observation the measured system can not be considered as separate from the observed phenomena. The intervention of the measuring instrument destroys all causal connection between the state of the system before and after the measurement; this explains why one cannot in general predict with certainty in what state the system will be found after the measurement.

 

[Fredrik]

This "inability to repeat measurements" is in my opinion better described as an inability to prepare a state with the desired properties (or as the non-existence of such a state in the mathematical part of QM). Since you measure the momentum by measuring the position, you can measure both with an accuracy that's only limited by the size of the detector.

 

[Naty1]

This "inability to repeat measurements" is in my opinion better described as an inability to prepare a state with the desired properties (or as the non-existence of such a state in the mathematical part of QM).

I'm surprised, if I understand what you posted: Zapper's blog which I quoted above seems to me a bit different:

Where the HUP comes into play is that if you then repeat the same sequence of arbitrarily precise measurements on a large numbers of identically prepared particles (i.e. particles with the same wave function, or equivalently particles sampled from the same probability distribution), you will find that your momentum measurements are not all identical, but rather form a probability distribution of possible values for the momentum.

But I haven't quite been able to figure out exactly what "prepare a state" means which Messiah in QUANTUM MECHANICS also mentions but doesn't explain. Where is Zapper?

 

[Fredrik]

It's not different. The "desired" properties are precisely those properties that would ensure that the results of all the momentum measurements (on different members of an ensemble of identically prepared systems) are essentially the same, and that the results of all the position measurements (on different members of the same ensemble) are essentially the same. ZZ's statement explains what my statement means.

I just don't like the phrase "repeated measurements", because it sounds like it might be referring to something you do repeatedly to the same particle (without re-preparing it between measurements) rather than to the members of an ensemble of identically prepared particles.

To prepare a state is just to bring a particle on which we intend to do a measurement to the measuring device. Different ways of doing that may give us different average results. Two ways of doing it (two preparation procedures) are considered equivalent if no series of measurements can distinguish between them (i.e. if they give us the same wavefunction, or more generally, the same state operator/density matrix). These equivalence classes are often called "states".

 

[Naty1]

Fredrik: thanks for the assistance...I have a bit more thinking to do, but I "get" the last two of your three paragraphs...

"It's not different". well, THAT's a relief! maybe wording semantics got in the way...

Your explanation of "to prepare a state" clarifies what that means...I sure do not like that terminology, but maybe that's just me...

 

[Naty1]

eaglelake:

I just read your early post...

Bouncing a photon off an atom tells us nothing about any uncertainties. We must bounce many identically prepared photons off like atoms in order to get the statistical distributions of atomic position measurements and atomic momentum measurements. What we call "uncertainty" is a property of a statistical distribution. You cannot determine an uncertainty from a single measurement. I hope this helps.

BRAVO!...concise, well stated...

 

[Roo2]

I've always liked how ZapperZ explains this in is blog http://physicsandphysicists.blogspot.com/2006/11/misconception-of-heisenberg-uncertainty.html" .

Is there a peer-reviewed article which presents this argument (that the limit of accuracy for the measurement of a single electron is technology rather than HUP)? It makes sense to me intuitively and mirrors what I was taught, but I'd like to see it addressed formally rather than via blog.

 

[CJames]

Is there something wrong with the link Fredrik posted earlier?

http://www.kevinaylward.co.uk/qm/ballentine_ensemble_interpretation_1970.pdf

Page 365

I don't mean this to be sarcastic if it comes across that way.

 

[abaio]

Great informoation. Thanks. It would have been awsome to be there when the Quantum giants were discussing and racing to find new discoveries in the new mysterious quantum world. Bohr vs. Einstein was a great duel...kind of that like Edison vs. Tesla.

 

[Roo2]

Is there something wrong with the link Fredrik posted earlier?

http://www.kevinaylward.co.uk/qm/ballentine_ensemble_interpretation_1970.pdf

Page 365

I don't mean this to be sarcastic if it comes across that way.

Thanks! I was skimming this thread at work and didn't notice the link. That's exactly what I was looking for. I was actually amused to find this thread; I had a discussion with my room mate a couple weeks ago about HUP. Both of us are college undergrads, and our experience with quantum mechanics is limited to one semester of physical chemistry apiece (at different universities). I mentioned that I had been taught that HUP applied to sets of measurements of p and r and didn't apply to the position and momentum of a single particle. He replied that since the particle is itself a wave, and a wave has no defined absolute position, then HUP applies to a single particle. I don't have the physics background to respond to that, so I let it be. That paper is clearing up the picture for me. Forgive me for asking again for references, but I don't have the physics education to know off the bat... have there been any revisions to the interpretation of the statistical model in regard to a single particle measurement in the 40 years since this paper was published?

 

[fuesiker]

The most intuitive way, imho, to view the uncertainty principle is to recognize that the wavefunction of some system in momentum space is the Fourier transform of the wavefunction of that same system in position space. You can arrive at this by honoring the fact that the momentum operator is the propagator of translation, and that is how you can relate both wavefunctions to each other to arrive at the equation showing they are related by the Fourier transform.

Now that we are convinced of this, we can look at the properties of the Fourier transform. If a function K is the Fourier transform of a function F, then the sharper F is, the broader is K.

The uncertainty principle has nothing to do with observation. It does not arise due to observation. It is the nature of things that there are non-commuting observables (time and energy, momentum and position) that cannot be determined simultaneously to arbitrary precision. We find this unintuitive because in our universe, Planck's constant is way too small for quantum effects to have become intuitive to our senses, just like special relativisitic effects are also counter-intuitive to our senses since c is so big in our world. Imagine if we lived in a world where c is 30 mph ;)

 

[Dickfore]

@OP:

The canonical commutation relation:
$$[\hat{x}_{i}, \hat{p}_{k}] = i \, \hbar \, \delta_{i k}$$

EDIT:
How to derive these relations? You need to identify momentum as the generator of translations in space (according to the correspondence principle and the fact that the physical quantity conserved due to homogeneity of space is linear momentum) and it acts on states that are eigenstates of position according to:
$$\hat{T}(\mathbf{a}) \, \left|\mathbf{x}\right\rangle = \left|\mathbf{x} + \mathbf{a}\right\rangle, \; \hat{T}(\mathbf{a}) = \exp{\left(-\frac{i}{\hbar} \, \mathbf{a} \cdot \hat{\mathbf{p}}\right)}$$

 

[ZealScience]

I heard that these states of momentum and position are orthogonal in Hilbert space. Probably mathematics thinks that it is intuitive, but not us. Because in experiments that can be seen with naked eyes have no violation to classical mechanics which is described by definite phases...

 

[Dickfore]

$$\begin{array}{l} \hat{x}_{i} \, \hat{T}(\mathbf{a}) \, \left|\mathbf{x}\right\rangle = (x_{i} + a_{i}) \, \left|\mathbf{x} + \mathbf{a}\right\rangle \\ \hat{T}(\mathbf{a}) \, \hat{x}_{i} \, \left|\mathbf{x}\right\rangle = x_{i} \, \left|\mathbf{x} + \mathbf{a}\right\rangle \end{array} \Rightarrow \left[\hat{x}_{i}, \hat{T}(\mathbf{a})\right] = a_{i} \, \hat{T}(\mathbf{a})$$

Expanding to linear power in the translation vector $\mathbf{a}$, we get:
$$-\frac{i}{\hbar} \, a_{k} \, \left[\hat{x}_{i}, \hat{p}_{k}\right] = a_{i} = \delta_{i k} \, a_{k}$$
Since this has to hold for arbitrary components $a_{k}$, we must have the commutation relations to hold.

After you have a commutation relation between two operators, you can apply the same math that you mentioned in the OP (Cauchy-Schwartz ineqality) to derive an Uncertainty relation.

 

[Dickfore]

I heard that these states of momentum and position are orthogonal in Hilbert space.

They are not. For example:
$$\langle x | p \rangle = \frac{1}{(2\pi \hbar)^{1/2}} \, \exp{\left(\frac{i}{\hbar} \, p \, x\right)} \neq 0$$

 

[ZealScience]

They are not. For example:
$$\langle x | p \rangle = \frac{1}{(2\pi \hbar)^{1/2}} \, \exp{\left(\frac{i}{\hbar} \, p \, x\right)} \neq 0$$

Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.

 

[Dickfore]

Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.

FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.

 

[ZealScience]

FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.

Accidents can happen... it is not deterministic~~~

 

[Fredrik]

FFFFFFFFFFFFF

I pressed the page back button on my keyboard and my whole response is gone. I hate this glitch on PF.

I have found that you can usually get it back if you press the page forward button and answer yes to the question if you want to send the information again.

 

[Dickfore]

I have found that you can usually get it back if you press the page forward button and answer yes to the question if you want to send the information again.

For me the text gets erased.

 

[atyy]

Good post nkadambi, but I must point out an inaccuracy in what you said. I didn't really understand this myself until recently. It is possible to measure position and momentum simultaneously. In fact, we often measure the momentum by measuring the position and interpreting the result as a momentum measurement. (Check out figure 3 in this pdf).

If I remember correctly, after measuring position, the particle collapses to an eigenstate of position, then spreads out due to time evolution. After measuring momentum, it collapses to an eigenstate of momentum, then remains in the same state since it is an eigenstate of the free particle Hamiltonian. What state does it collapse to after position and momentum are simultaneously measured?

 

[Fredrik]

If I remember correctly, after measuring position, the particle collapses to an eigenstate of position, then spreads out due to time evolution. After measuring momentum, it collapses to an eigenstate of momentum, then remains in the same state since it is an eigenstate of the free particle Hamiltonian. What state does it collapse to after position and momentum are simultaneously measured?

In this example, the particle is absorbed by the detector, so no new state is prepared. If it had been the kind of detector that let's the particle pass through it, the new state would have been one with a sharply defined position (the wavefunction will be close to zero far from the detector).

After measuring momentum, it collapses to an eigenstate of momentum

This is what I was taught as well, but I wonder if it really makes sense. Even if we disregard all the difficulties associated with the fact that an eigenfunction of $-i\,\nabla$ isn't square-integrable, and interpret the statement as "after a momentum measurement, the wavefunction will have a sharply defined momentum", it still looks false to me. I don't know much about how things are measured, but it seems to me that momentum is always measured by measuring the position instead. It's possible that in most cases, the position measurement is inexact enough to give us something like exp(-x²) as the new wavefunction, so that both the wavefunction and its Fourier transform have a single peak, with widths of the same order of magnitude in units such that $\hbar=1$.

 

[atyy]

This is what I was taught as well, but I wonder if it really makes sense. Even if we disregard all the difficulties associated with the fact that an eigenfunction of $-i\,\nabla$ isn't square-integrable, and interpret the statement as "after a momentum measurement, the wavefunction will have a sharply defined momentum", it still looks false to me. I don't know much about how things are measured, but it seems to me that momentum is always measured by measuring the position instead. It's possible that in most cases, the position measurement is inexact enough to give us something like exp(-x²) as the new wavefunction, so that both the wavefunction and its Fourier transform have a single peak, with widths of the same order of magnitude in units such that $\hbar=1$.

I haven't really thought this through myself, and am just going to ask questions as they come to mind (but OP please stop me if this is hijacking!). How about in the relativistic case, say like Compton scattering? Isn't momentum there measured without measuring position? OTOH, there is no "sensible" position operator in relativistic QFT, so maybe that's different?

 

[Fredrik]

Can you specify the eigenvectors of position and momentum? I just started learning it and I am not familiar with it.

Neither of those operators have eigenvectors in the semi-inner product space of square-integrable functions from $\mathbb R$ into $\mathbb C$. If we view this space as a subspace of the vector space of all functions from $\mathbb R$ into $\mathbb C$, then the functions $u_p$ defined by $u_p(x)=e^{ipx}$ for all x are eigenfunctions (with eigenvalue p) of the momentum operator. Note that they are not square-integrable. The position operator doesn't have any eigenfunctions in this space either, but there's a trick. The "functions" $v_x$ such that $v_x(y)=\delta(y-x)$ for all y, where $\delta$ is the Dirac delta, can be thought of as "eigenfunctions" (with "eigenvalue" x) of the position operator, even though they aren't really functions.

 

[Fredrik]

How about in the relativistic case, say like Compton scattering? Isn't momentum there measured without measuring position?

I don't think so. The way I see it, a measuring device is just a device that produces a signal (that can be approximately described as classical) that informs us that an interaction has taken place. The location of the device (or the location of the relevant component of it) can always be interpreted as the result of a position measurement. So it's not possible to measure anything without measuring the position of the particle(s) that participated in the interaction that produced the signal.

Hm, I suppose it's possible that those particles aren't the same ones as the one we're trying to measure the momentum of.

 

[Dickfore]

Regarding measurement and falling back into the eigenstate of the observable corresponding to the observed eigenvalue, I have doubts

If you read section 7 in "Quantum Mechanics vol. 3" (the wave function and measurements) by Landau, Lifgarbagez, they discuss in the 12th paragraph the possibility that the wave function in which the electron drops after the measurement is not necessarily an eigenfunction of the observable being measured.